Quantum Mechanics - Angular Momentum probability

  • Thread starter Tangent87
  • Start date
  • #1
Tangent87
148
0
See the attachment for the question.

I'm stuck on the bit highlighted in yellow (have done the rest of the question).

I'm not sure if it's really trivial or more involved. Do I have to use the fact that U(pi/2)|j m> are eigenvectors of J1 with eigenvalue m somehow? Why is the answer not just:
|<j m'| J1 |j m>|2?

Thanks
 

Attachments

  • pqm2003.jpg
    pqm2003.jpg
    35.1 KB · Views: 607

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,566
2,204
You don't generally apply the operator when you're making a measurement. Instead, what you do is expand the state of the system in terms of the eigenstates of the observable

[tex]\vert j m \rangle = \sum_{m' = -j}^j c_{m'} \vert j m' \rangle_{J_1}[/tex]

(where [itex]\vert j m' \rangle_{J_1}[/itex] are the eigenstates of J1) and then pick off the coefficients (taking advantage of the orthonormality of the eigenstates) and square them to calculate the probabilities.
 
  • #3
Tangent87
148
0
You don't generally apply the operator when you're making a measurement. Instead, what you do is expand the state of the system in terms of the eigenstates of the observable

[tex]\vert j m \rangle = \sum_{m' = -j}^j c_{m'} \vert j m' \rangle_{J_1}[/tex]

(where [itex]\vert j m' \rangle_{J_1}[/itex] are the eigenstates of J1) and then pick off the coefficients (taking advantage of the orthonormality of the eigenstates) and square them to calculate the probabilities.


Ah yes of course! I've done it now, cheers.
 

Suggested for: Quantum Mechanics - Angular Momentum probability

Replies
9
Views
187
Replies
3
Views
164
Replies
8
Views
205
Replies
22
Views
361
Replies
12
Views
250
  • Last Post
Replies
6
Views
180
  • Last Post
Replies
1
Views
559
Replies
2
Views
175
Top